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Musya8 [376]
3 years ago
6

What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP

a at a strain of 0.003
Engineering
2 answers:
Temka [501]3 years ago
4 0

Answer:

U = 0.75 MPa

Therefore, the modulus of resilience for the tensile test specimen is 0.75MPa

Explanation:

The Modulus of resilience can be derived from the area under the elastic portion of the stress-strain diagram. It can be written mathematically as;

U = 1/2 × yield strength × yield strain.

Where U is the modulus of resilience.

Given;

Yield strength = 500MPa

Yield strain = 0.003

Substituting the given values we have;

U = 1/2 × 500MPa × 0.003

U = 0.75 MPa

Therefore, the modulus of resilience for the tensile test specimen is 0.75MPa

ella [17]3 years ago
3 0

Answer:

The modulus of resilience is 166.67 MPa

Explanation:

Modulus of resilience is given by yield strength ÷ strain

Yield strength = 500 MPa

Strain = 0.003

Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa

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7 0
3 years ago
Each cout statement has a syntax error. Type the first cout statement, and press Run to observe the error message. Fix the error
saul85 [17]

Answer:

1. cout << "Num: " << songNum << endl;

2. cout << songNum << endl;

3. cout << songNum <<" songs" << endl;

Explanation:

//Full Code

#include <iostream>

using namespace std;

int main ()

{

int songNum;

songNum = 5;

cout << "Num: " << songNum << endl;

cout << songNum << endl;

cout << songNum <<" songs" << endl;

return 0;

}

1. The error in the first cout statement is that variable songnum is not declared.

C++ is a case sensitive programme language; it treats upper case and lower case characters differently.

Variable songNum was declared; not songnum.

2. Cout us used to print a Variable that has already been declared.

The error arises in int songNum in the second cout statement.

3. When printing more than one variables or values, they must be separated with <<

4 0
3 years ago
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
Bad White [126]

Answer:

the net work per cycle \mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume V_2 = 0.16 V_1

the crankshaft N rotates at a speed of  2400 RPM.

At the beginning of the compression , temperature T_1 = 60 F = 519.67 R    

and;

Otto cycle with a pressure =  14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

V_1-V_2 = \dfrac{\pi}{4}D^2L

V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)

V_1-0.16V_1= 36.55714291

0.84 V_1 =36.55714291

V_1 =\dfrac{36.55714291}{0.84 }

V_1 =43.52040823 \ in^3 \\ \\  V_1 = 43.52 \ in^3

V_1 = 0.02518 \ ft^3

the mass in air ( lb) can be determined by using the formula:

m = \dfrac{P_1V_1}{RT}

where;

R = 53.3533 ft.lbf/lb.R°

m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R  \times 519 .67 ^0 R}

m = 0.0018962 lb

From the tables  of ideal gas properties at Temperature 519.67 R

v_{r1} =158.58

u_1 = 88.62 Btu/lb

At state of volume 2; the relative volume can be determined as:

v_{r2} = v_{r1}  \times \dfrac{V_2}{V_1}

v_{r2} = 158.58 \times 0.16

v_{r2} = 25.3728

The specific energy u_2 at v_{r2} = 25.3728 is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

v_{r3} = 0.1828

u_3 = 1098 \ Btu/lb

To determine the relative volume at state 4; we have:

v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}

v_{r4} =0.1828 \times \dfrac{1}{0.16}

v_{r4} =1.1425

The specific energy u_4 at v_{r4} =1.1425 is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

W_{net} = Heat  \ supplied - Heat  \ rejected

W_{net} = m(u_3-u_2)-m(u_4 - u_1)

W_{net} = m(u_3-u_2- u_4 + u_1)

W_{net} = m(1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (410.08)

\mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the  four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

W = 4 \times N'  \times W_{net

where ;

N' = \dfrac{2400}{2}

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec ×  0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0
2 years ago
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